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zzr0ck3r
 one year ago
Good methods for finding eigen values?
\(M =
\begin{array}{cc}
0&1&1&1&1& \\
1&0&1&1&1 \\
1&1&0&1&1 \\
1&1&1&0&1\\
1&1&1&1&0
\end{array}\)
zzr0ck3r
 one year ago
Good methods for finding eigen values? \(M = \begin{array}{cc} 0&1&1&1&1& \\ 1&0&1&1&1 \\ 1&1&0&1&1 \\ 1&1&1&0&1\\ 1&1&1&1&0 \end{array}\)

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zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1@freckles where you go ;(

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0\[\det(A \lambda I) =0\] and then use cofactor expansion like crazy

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I said good methods :)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0row operations first? idk that may destroy the determinant...

freckles
 one year ago
Best ResponseYou've already chosen the best response.2lol well ... I was kind of trying to see if could find a pattern but I didn't see a pattern like you know M_n is a n by n matrix with 0's on the diagonal and 1's everywhere else but it was totally not a good method

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and not just a bad method but it didn't seem to work like that

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0does this whole matrix break one of the rules for row operations? Don't we need a 1 at the top left corner of the first row?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0row swaps... so it does not break the first rule

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I would switch the last one with the first one r_1<>r_5

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0swap r_2 <> r_4 , then r_3 <> r_4 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2actually I do see a pattern

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so there must be some kinda of shortcut for M_n where M_n is a n by n matrix with 0's a long the diagonal and 1's elsewhere

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I want to make a conjecture \[\text{ egienvalues }(M_n)=n1,1 \text{ (multiplicity } (n1) \text{ )}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Now I know this isn't a method but it might help to come up with something from this if it is even true

freckles
 one year ago
Best ResponseYou've already chosen the best response.2like we know the trace of M is suppose to be the sum of the eigenvalues

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[0=\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I don't know a quick way to find the determinant but the product of the eigenvalues equals the determinant

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1ill play with this too

freckles
 one year ago
Best ResponseYou've already chosen the best response.2another cute thing \[M_n=[1]_nI\] where 1_n is a n by n matrix only made of 1's and I is the identity matrix don't know if this helps yet

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you know I'm not totally sure yet but it seems to be: in this case anyways \[\text{ eigenvalues(}M_n\text{)}=\text{ egivenvalues(}[1]_n\text{)}\text{ egienvalues(}I\text{)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the eigenvalues of I are 1 (n amount of times) and the eigenvalues of [1]_n are n, 0 (n1 amount of times)
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